Question

The following table lists molar concentrations of seven major ions in seawater. Using a density of 1.022 g/mL for seawater, convert the concentrations for the following two ions into molality. The sodium ion: The sulfate ion:

Answer 1

Let the mass of seawater be 1000 g or 1 kg.

Density of seawater = 1.022 g/mL

Density = Mass/Volume

1.022 = 1000/Volume

Volume of seawater = 978.47 mL

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Molarity of sodium ion = no. of moles of sodium ion/volume of seawater in litres

480.57 $\times$ 10^-3 = no. of moles of sodium ion/0.978

no. of moles of sodium ion = 0.469

Molality of sodium ion = no. of moles of sodium ion/volume of seawater in kgs

Molality of sodium ion = 0.469/1 = 0.469 m

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Molarity of sulfate ion = no. of moles of sulfate ion/volume of seawater in litres

28.93 $\times$ 10^-3 = no. of moles of sulfate ion/0.978

no. of moles of sulfate ion = 0.0283

Molality of sulfate ion = no. of moles of sulfate ion/volume of seawater in kgs

Molality of sulfate ion = 0.0283/1 = 0.0283 m

Answer 2

Volume of sea water = 1 L = 1000 ml

mass of sea water = 1000 ml x 1.022 g/ml = 1022 g

(a) the sodium ion

mass of sodium ion = 480.57 mM = 0.48057 mol/L

                                = 0.48057 mol x 23 g/mol

                                = 11.05311 g/L

mass of water in solution = 1022 - 11.05311 = 1010.9469 g

                                         = 1.0109469 kg

molality of sodium ions in sea water = 0.48057 mol/1.0109469 kg

                                                          = 0.4754 m

(b) the sulfate ion

mass of sulfate ion = 28.93 mM = 0.02893 mol/L

                                = 0.02893 mol x 96.06 g/mol

                                = 2.78 g/L

mass of water in solution = 1022 - 2.78 = 1019.22 g

                                         = 1.01922 kg

molality of sulfate ions in sea water = 0.02893 mol/0.101922 kg

                                                          = 0.0284 m